3.4.62 \(\int \sec ^3(\frac {\pi }{4}+\frac {x}{2}) \tan ^2(\frac {\pi }{4}+\frac {x}{2}) \, dx\) [362]

Optimal. Leaf size=76 \[ -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )-\frac {1}{4} \sec \left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right ) \]

[Out]

-1/4*arctanh(sin(1/4*Pi+1/2*x))-1/4*sec(1/4*Pi+1/2*x)*tan(1/4*Pi+1/2*x)+1/2*sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2
*x)

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Rubi [A]
time = 0.03, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2691, 3853, 3855} \begin {gather*} -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {x}{2}+\frac {\pi }{4}\right )\right )+\frac {1}{2} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec ^3\left (\frac {x}{2}+\frac {\pi }{4}\right )-\frac {1}{4} \tan \left (\frac {x}{2}+\frac {\pi }{4}\right ) \sec \left (\frac {x}{2}+\frac {\pi }{4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]

[Out]

-1/4*ArcTanh[Sin[Pi/4 + x/2]] - (Sec[Pi/4 + x/2]*Tan[Pi/4 + x/2])/4 + (Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2])/2

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx &=\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )-\frac {1}{4} \int \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx\\ &=-\frac {1}{4} \sec \left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )-\frac {1}{8} \int \csc \left (\frac {\pi }{4}-\frac {x}{2}\right ) \, dx\\ &=-\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )-\frac {1}{4} \sec \left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^3\left (\frac {\pi }{4}+\frac {x}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 74, normalized size = 0.97 \begin {gather*} -\frac {1}{4} \tanh ^{-1}\left (\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )-\frac {1}{4} \sec ^2\left (\frac {1}{4} (\pi +2 x)\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right )+\frac {1}{2} \sec ^4\left (\frac {1}{4} (\pi +2 x)\right ) \sin \left (\frac {\pi }{4}+\frac {x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[Pi/4 + x/2]^3*Tan[Pi/4 + x/2]^2,x]

[Out]

-1/4*ArcTanh[Sin[Pi/4 + x/2]] - (Sec[(Pi + 2*x)/4]^2*Sin[Pi/4 + x/2])/4 + (Sec[(Pi + 2*x)/4]^4*Sin[Pi/4 + x/2]
)/2

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Maple [A]
time = 0.38, size = 76, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\)	\(76\)
default	\(\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{2 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{4}}+\frac {\sin ^{3}\left (\frac {\pi }{4}+\frac {x}{2}\right )}{4 \cos \left (\frac {\pi }{4}+\frac {x}{2}\right )^{2}}+\frac {\sin \left (\frac {\pi }{4}+\frac {x}{2}\right )}{4}-\frac {\ln \left (\sec \left (\frac {\pi }{4}+\frac {x}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {x}{2}\right )\right )}{4}\)	\(76\)
risch	\(\frac {i \left (-\left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {7 i x}{2}}+7 \left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {5 i x}{2}}+7 \left (-1\right )^{\frac {3}{4}} {\mathrm e}^{\frac {3 i x}{2}}-\left (-1\right )^{\frac {1}{4}} {\mathrm e}^{\frac {i x}{2}}\right )}{2 \left (i {\mathrm e}^{i x}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}-i\right )}{4}-\frac {\ln \left ({\mathrm e}^{\frac {i \left (\pi +2 x \right )}{4}}+i\right )}{4}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(1/4*Pi+1/2*x)^3*tan(1/4*Pi+1/2*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*sin(1/4*Pi+1/2*x)^3/cos(1/4*Pi+1/2*x)^4+1/4*sin(1/4*Pi+1/2*x)^3/cos(1/4*Pi+1/2*x)^2+1/4*sin(1/4*Pi+1/2*x)-
1/4*ln(sec(1/4*Pi+1/2*x)+tan(1/4*Pi+1/2*x))

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Maxima [A]
time = 1.71, size = 74, normalized size = 0.97 \begin {gather*} \frac {\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{3} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} - 2 \, \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} + 1\right )}} - \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + \frac {1}{8} \, \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="maxima")

[Out]

1/4*(sin(1/4*pi + 1/2*x)^3 + sin(1/4*pi + 1/2*x))/(sin(1/4*pi + 1/2*x)^4 - 2*sin(1/4*pi + 1/2*x)^2 + 1) - 1/8*
log(sin(1/4*pi + 1/2*x) + 1) + 1/8*log(sin(1/4*pi + 1/2*x) - 1)

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Fricas [A]
time = 1.18, size = 82, normalized size = 1.08 \begin {gather*} -\frac {\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) - \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4} \log \left (-\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 1\right ) + 2 \, {\left (\cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{2} - 2\right )} \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{8 \, \cos \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="fricas")

[Out]

-1/8*(cos(1/4*pi + 1/2*x)^4*log(sin(1/4*pi + 1/2*x) + 1) - cos(1/4*pi + 1/2*x)^4*log(-sin(1/4*pi + 1/2*x) + 1)
+ 2*(cos(1/4*pi + 1/2*x)^2 - 2)*sin(1/4*pi + 1/2*x))/cos(1/4*pi + 1/2*x)^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \tan ^{2}{\left (\frac {x}{2} + \frac {\pi }{4} \right )} \sec ^{3}{\left (\frac {x}{2} + \frac {\pi }{4} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)**3*tan(1/4*pi+1/2*x)**2,x)

[Out]

Integral(tan(x/2 + pi/4)**2*sec(x/2 + pi/4)**3, x)

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Giac [A]
time = 1.61, size = 95, normalized size = 1.25 \begin {gather*} \frac {\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )}{4 \, {\left ({\left (\frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )\right )}^{2} - 4\right )}} - \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) + 2 \right |}\right ) + \frac {1}{16} \, \log \left ({\left | \frac {1}{\sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right )} + \sin \left (\frac {1}{4} \, \pi + \frac {1}{2} \, x\right ) - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(1/4*pi+1/2*x)^3*tan(1/4*pi+1/2*x)^2,x, algorithm="giac")

[Out]

1/4*(1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1/2*x))/((1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1/2*x))^2 - 4) - 1/16
*log(abs(1/sin(1/4*pi + 1/2*x) + sin(1/4*pi + 1/2*x) + 2)) + 1/16*log(abs(1/sin(1/4*pi + 1/2*x) + sin(1/4*pi +
1/2*x) - 2))

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Mupad [B]
time = 6.38, size = 75, normalized size = 0.99 \begin {gather*} \frac {2\,\left (\frac {{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^7}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^5}{4}+\frac {7\,{\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^3}{4}+\frac {\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}{4}\right )}{{\left ({\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )}^2-1\right )}^4}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {\Pi }{8}+\frac {x}{4}\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(Pi/4 + x/2)^2/cos(Pi/4 + x/2)^3,x)

[Out]

(2*(tan(Pi/8 + x/4)/4 + (7*tan(Pi/8 + x/4)^3)/4 + (7*tan(Pi/8 + x/4)^5)/4 + tan(Pi/8 + x/4)^7/4))/(tan(Pi/8 +
x/4)^2 - 1)^4 - atanh(tan(Pi/8 + x/4))/2

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